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A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.9 · ~6 min read

Complex numbers

Arithmetic with complex numbers, conjugates, modulus and argument, the Argand diagram, square roots of a complex number, conjugate roots of real polynomials, the forms r(cos θ + i sin θ) and reiθ, and loci and regions.

i and x + iyConjugateDivisionModulus & argumentArgand diagramSquare rootsConjugate-pair rootsPolar & exponential form
RSby Riyaz Sir
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • i² = −1. Treat x + iy like an algebraic expression, replacing i² by −1.
  • To divide, multiply top and bottom by the conjugate of the bottom; the denominator becomes real.
  • z = r(cos θ + i sin θ) = reiθ: multiplying multiplies the moduli and adds the arguments; dividing divides the moduli and subtracts the arguments.
  • Loci: ǀz − aǀ = k is a circle; ǀz − aǀ = ǀz − bǀ is a perpendicular bisector; arg(z − a) = α is a half-line from a.

By the end you can

  • Add, subtract, multiply and divide complex numbers in the form x + iy, and use conjugates.
  • Find the modulus and argument, using −π < θ ≤ π, and represent numbers on an Argand diagram.
  • Find square roots of a complex number and solve polynomial equations with complex roots.
  • Use the forms r(cos θ + i sin θ) and reiθ to multiply and divide.
  • Sketch loci and regions defined by modulus and argument.

1Cartesian form

Arithmetic

Conjugate — of z = x + iy is z* = x − iy. zz* = x² + y² is real.

  • Equal complex numbers: real parts equal and imaginary parts equal. This gives two equations from one.
Worked example

Express (3 + 4i)/(1 − 2i) in the form x + iy.

  1. multiply by (1 + 2i)/(1 + 2i) M1
  2. (3 + 6i + 4i − 8)/(1 + 4) = (−5 + 10i)/5 A1
  3. −1 + 2i A1
Worked example

Find z such that 2z + iz* = 3 + 6i.

  1. z = x + iy: 2x + 2iy + ix + y = 3 + 6i M1
  2. real: 2x + y = 3; imaginary: x + 2y = 6 M1
  3. x = 0, y = 3: z = 3i A1
Worked example

Find the square roots of 5 − 12i.

  1. (a + ib)² = 5 − 12i ⇒ a² − b² = 5, 2ab = −12 M1
  2. b = −6/a ⇒ a⁴ − 5a² − 36 = 0 ⇒ a² = 9 M1
  3. ±(3 − 2i) A1 A1

Polynomial roots

  • Real coefficients ⇒ non-real roots come in conjugate pairs. If 1 + 2i is a root, so is 1 − 2i, and (z − 1)² + 4 = z² − 2z + 5 is a factor.
Worked example

Given that 1 + 2i is a root of z³ − 4z² + 9z − 10 = 0, find the other roots.

  1. 1 − 2i is also a root B1
  2. quadratic factor z² − 2z + 5 M1 A1
  3. z³ − 4z² + 9z − 10 = (z² − 2z + 5)(z − 2), so the roots are 2, 1 ± 2i A1
Common mistakeCalculator answers without working. You may check on a calculator, but division and roots must be shown algebraically.

2Modulus and argument

The Argand diagram

ReIm-3-2-1123-2-1120z = −1 + √3 iz* = −1 − √3 iarg z = 2π/3|z| = 2z = r(cos θ + i sin θ)= re^(iθ) = r cis θr = 2, θ = 2π/3
z = x + iy is the point (x, y); r is its distance from O and θ its angle from the positive real axis.
ǀzǀ = √(x² + y²) ; arg z = θ, with tan θ = y/x and −π < θ ≤ π
always sketch to choose the quadrant
  • Quadrants: for −1 + √3 i (second quadrant), tan⁻¹(√3/1) = π/3, so arg z = π − π/3 = 2π/3, and ǀzǀ = 2.
z = r(cos θ + i sin θ) = reiθ
polar and exponential forms
ǀz₁z₂ǀ = ǀz₁ǀǀz₂ǀ, arg(z₁z₂) = arg z₁ + arg z₂ ; ǀz₁/z₂ǀ = ǀz₁ǀ/ǀz₂ǀ, arg(z₁/z₂) = arg z₁ − arg z₂ LEARN IT
adjust by 2π to stay in (−π, π]
Worked example

z = 2eiπ/3 and w = √2 e−iπ/4. Find zw and z/w in the form reiθ.

  1. zw = 2√2 ei(π/3 − π/4) = 2√2 eiπ/12 M1 A1
  2. z/w = √2 ei(π/3 + π/4) = √2 e7iπ/12 A1
Common mistakeArguments from tan⁻¹(y/x) alone. For −1 − i, tan⁻¹(1) = π/4 but the argument is −3π/4. The 2024 Paper 3 report flagged quadrant errors and degree answers.

3Loci and regions

Three standard loci

ReImǀz − (2 + i)ǀ ≤ 2disc, centre 2 + i, radius 2ReImǀz − 1ǀ = ǀz − 3iǀperpendicular bisectorReImarg(z − 1 − i) = π/4half-line from 1 + iThree standard loci on an Argand diagramFixed distance · equal distances · fixed angle
A circle (fixed distance), a perpendicular bisector (equal distances) and a half-line (fixed angle).
  • ǀz − (2 + i)ǀ ≤ 2: all points within 2 of 2 + i: a disc. Strict "<" uses a dashed boundary.
  • ǀz − 1ǀ = ǀz − 3iǀ: points equidistant from 1 and 3i: the perpendicular bisector of the segment joining them.
  • arg(z − 1 − i) = π/4: a half-line starting at 1 + i (not including that point) at 45° to the horizontal.
Worked example

Find the Cartesian equation of ǀz − 1ǀ = ǀz − 3iǀ.

  1. (x − 1)² + y² = x² + (y − 3)² M1
  2. −2x + 1 = −6y + 9 A1
  3. x − 3y + 4 = 0 (y = (x + 4)/3) A1
Worked example

z lies on the circle ǀz − 3 − 4iǀ = 2. Find the least and greatest values of ǀzǀ.

  1. the centre 3 + 4i is 5 from O M1
  2. least = 5 − 2 = 3, greatest = 5 + 2 = 7 A1 A1
Common mistakeDrawing arg(z − a) = α as a full line. It is a half-line from a, and a is excluded.
Linking questionModulus and argument link to 2D vectors; the factor theorem used here is in Algebra; and the trig values come from Further trigonometry.

Test yourself — cover the answers first

Quick check

  1. Simplify i³.
  2. Find (2 + i)(2 − i).
  3. Find ǀ−5 + 12iǀ.
  4. Find arg(−2), with −π < θ ≤ π.
  5. If 3 − i is a root of a real cubic, name another root.

Exam-style

  1. Express (1 + i)⁴ in the form x + iy using the exponential form.
  2. Find the modulus and argument of (1 + i√3)/(1 − i).
  3. Shade the region ǀz − 2ǀ ≤ 2 and Re z ≥ 1: describe it.
Show answers
  1. −i
  2. 5
  3. 13
  4. π
  5. 3 + i
  6. (√2 eiπ/4)⁴ = 4eiπ = −4
  7. modulus 2/√2 = √2; argument π/3 + π/4 = 7π/12
  8. inside the disc centre 2, radius 2, to the right of the line x = 1
Practise this chapter
A Level practice papers
Next chapterForces and equilibrium →
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.